Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x-8y &= 6 \\ 8x-2y &= -3\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $8x = 2y-3$ Divide both sides by $8$ to isolate $x$ $x = {\dfrac{1}{4}y - \dfrac{3}{8}}$ Substitute this expression for $x$ in the first equation. $8({\dfrac{1}{4}y - \dfrac{3}{8}}) - 8y = 6$ $2y - 3 - 8y = 6$ Simplify by combining terms, then solve for $y$ $-6y - 3 = 6$ $-6y = 9$ $y = -\dfrac{3}{2}$ Substitute $-\dfrac{3}{2}$ for $y$ in the top equation. $8x-8( -\dfrac{3}{2}) = 6$ $8x+12 = 6$ $8x = -6$ $x = -\dfrac{3}{4}$ The solution is $\enspace x = -\dfrac{3}{4}, \enspace y = -\dfrac{3}{2}$.